How to Properly Compare Two Integers in Java

Is it safe to compare two `Integer` values with `==` in Java?

No, it's not the right way to compare the Integer objects. You should use Integer.equals() or Integer.compareTo() method.

By default JVM will cache the Integer values from [-128, 127] range (see java.lang.Integer.IntegerCache.high property) but other values won't be cached:

Integer x = 5000;
Integer y = 5000;
System.out.println(x == y); // false

Unboxing to int or calling Integer.intValue() will create an int primitive that can be safely compared with == operator. However unboxing a null will result in NullPointerException.

How do I compare two Integers?

This is what the equals method does:

public boolean equals(Object obj) {
    if (obj instanceof Integer) {
        return value == ((Integer)obj).intValue();
    }
    return false;
}

As you can see, there's no hash code calculation, but there are a few other operations taking place there. Although x.intValue() == y.intValue() might be slightly faster, you're getting into micro-optimization territory there. Plus the compiler might optimize the equals() call anyway, though I don't know that for certain.

I generally would use the primitive int, but if I had to use Integer, I would stick with equals().

How do I compare two Integers?

This is what the equals method does:

public boolean equals(Object obj) {
    if (obj instanceof Integer) {
        return value == ((Integer)obj).intValue();
    }
    return false;
}

As you can see, there's no hash code calculation, but there are a few other operations taking place there. Although x.intValue() == y.intValue() might be slightly faster, you're getting into micro-optimization territory there. Plus the compiler might optimize the equals() call anyway, though I don't know that for certain.

I generally would use the primitive int, but if I had to use Integer, I would stick with equals().

Java Compare 2 integers with equals or ==?

int is a primitive. You can use the wrapper Integer like

Integer first_int = 1;
Integer second_int = 1;
if(first_int.equals(second_int)){ // <-- Integer is a wrapper.

or you can compare by value (since it is a primitive type) like

int first_int = 1;
int second_int = 1;
if(first_int == second_int){ // <-- int is a primitive.

JLS-4.1. The Kinds of Types and Values says (in part)

There are two kinds of types in the Java programming language: primitive types (§4.2) and reference types (§4.3). There are, correspondingly, two kinds of data values that can be stored in variables, passed as arguments, returned by methods, and operated on: primitive values (§4.2) and reference values (§4.3).

Comparing two Integer Wrapper classes

Using the new keyword always creates two different instances. So the following is always true:

new Integer(10) != new Integer(10)

hence the first line printing "false".

Then:

i++;

hides the unboxing and boxing. It is equivalent to:

i = Integer.valueOf(i.intValue() + 1);

As described in the Javadoc of Integer.valueOf, values from -128 to 127 (at least) are cached: you are getting back the cached instance of Integer.valueOf(11) for both i++ and i1++, hence the second line printing "true".

Comparing Integer objects

For reference types, == checks whether the references are equal, i.e. whether they point to the same object.

For primitive types, == checks whether the values are equal.

java.lang.Integer is a reference type. int is a primitive type.

Edit: If one operand is of primitive type, and the other of a reference type that unboxes to a suitable primitive type, == will compare values, not references.

---

Related Topics